Given a byte array, which is an encoding of characters. Here is the rule:
    a. If the first bit of a byte is 0, that byte is a one-byte character
    b. If the first bit of a byte is 1, that byte and its following byte
       together stand for a two-byte character

Now implement a function to decide if the last character is a one-byte character or a two-byte character. You must scan the byte array from the end to the start. Otherwise it is very trivial.

思路:

B1....Bn.
从后往前：
   如果Bn是1开头，那么2字节结尾。
   如果Bn是0开头，看Bn-1:
      如果是0开头，无论Bn-1是自己管自己还是Bn-2的附属，都不影响Bn,所以1字节结尾。
      如果是1开头就要继续看前面，记录下已经看到的1开头的个数C.

直到找到一个0开头的或者到头了就结束。

如果C偶数，那么1字节结尾。
如果C为奇数，那么2字节结尾。

例如:

Bn
1 -> 2-byte end
0 -> need to look at Bn-1:

Bn-1 Bn
0    0 -> 1-byte end
1    0 -> need to look at Bn-2:

Bn-2 Bn-1 Bn
0    1    0 -> 2-byte end
1    1    0 -> need to look at Bn-3:

Bn-3 Bn-2 Bn-1 Bn
0    1    1    0 -> 1-byte end
1    1    1    0 -> need to look at Bn-4:
...


代码:

bool firstBitIsOne(unsigned char c) {
    return (c >> 7) & 1;
}

// return number of bytes in the last byte.
int lastByteCount(string s) {
    int n = s.length();
    if (n == 0) return 0;
    
    if (firstBitIsOne(s[n-1])) return 2;
    
    int i = n-2;
    for (; i >= 0 && firstBitIsOne(s[i]); -- i) {}
    
    if ((n - i) & 1) return 2; // odd number of 1s.
    else return 1;
}


来自: http://www.mitbbs.com/article_t/JobHunting/32605453.html